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23

You can use a high vertical tube to store water in it (fill it from the bottom by pushing the water in) How much water can you store? It obviously depends on the pressure you apply to push it in. If you push harder, there will be more water stored. The tube is characterized not the amount of water, but by how easy it is to store the water. Its "capacity" ...


5

As drawn, the circuit, assuming ideal circuit elements, is problematic for the reason you've deduced (KVL gives a contradiction). One interpretation is that there is infinite large current for an infinitesimal time which instantaneously charges the capacitors to their final steady state voltages. To gain some insight, add a resistance $r$ in series with ...


5

A typical voltmeter contains an internal Ohmic resistor with known and very high resistance $R$ (called the "input resistance" or "input impedance"), and an extremely sensitive ammeter that measures the current through that resistor. When the voltmeter is connected in parallel across some circuit elements, then ideally the internal resistor has resistance ...


4

a. Immediately after the switch is closed, are either or both bulbs glowing? Explain. They will both glow as some current passes through them as the capacitor is charging. b. If both bulbs are glowing, which is brighter? Or are they equally bright? Explain. They are both equally bright, because an equal and opposite charge is flowing on to ...


4

Capacitance is "charge over voltage" – and one farad is "coulomb per volt" – because the capacity of capacitors (something that determines their "quality") is the ability to store a maximum charge on the plate ($+Q$ on one side, $-Q$ on the other side) given a fixed voltage. When you try to separate the charges, you unavoidably create electric fields ...


4

I think the question is quite interesting actually. When resistances are equal the voltage will divide itself equally because the resistances are coupled in series. Then we take each R to infinity and we assume that we do this to each resistance in the same manner. I think that the voltage across each resistance remains 5 Volts in this case when you take R ...


3

No, that circuit cannot exist in that regime. You are neglecting the internal resistance of the wires between the voltage source and the capacitors, and if the capacitors are discharged (in which case the voltage over them is zero) that's no longer a good approximation. You therefore need to insert a small resistance on either side of the voltage source, ...


3

We Use $C=Q/V$ because those were useful things to measure. It's often easy to forget, but many of the equations we use are chosen because the work, and because other equations didn't work. Never underestimate that part of the reality. We don't use "charge per unit volume" because that number is not constant. You can charge a capacitor up without ...


3

Remembering that resistance = $\frac V I$ work the resistance at $I=1$ and $I=2$. For the resistance to be constant the current-voltage characteristic must be a straight line and go through the origin. There is another parameter which is useful in some instances and that is called the incremental resistance $\frac {\Delta V} {\Delta I}$ which is related ...


3

I would try a more "dynamic" approach, where you actually have an RC circuit (https://en.wikipedia.org/wiki/RC_circuit ) - even a piece of wire has some non-zero capacitance. The typical time scale of transient processes in such circuits is $RC$. When $R\rightarrow \infty$, you have $RC\rightarrow \infty$, so you have to wait longer and longer for transients ...


3

Try redrawing it as so - should be easier to figure out. EDIT the top R2 in the second diagram should be R4.


3

From Kirchhoff's second law, the sum of all the voltages around a loop is equal to zero. That is, the sum of the voltages across the three elements of your circuit, R, L and C, must be equal to the time varying voltage from the source: $$V_R+V_L+V_C = V(t)$$ As $V_R=RI$, $V_L=L\frac{dI}{dt}$ and $V_C=\frac{Q}{C}$, we get your equation, which is correct: ...


3

So, we have series LCR circuit. $V$ is a constant voltage source. $L$, $C$, and $R$ represents the inductance, capacitance and resistance in the circuit respectively. A current $I$ flows through the circuit. Now, the current through each component is the same. So, the potential difference between each component added up together gives the emf $V$. ...


3

The fact that the cell has internal resistance & it acts as a source corrupts your argument. Your argument that connecting a resistance across one of the cells in parallel reduces the overall resistance of the circuit and hence expect the current to increase is wrong. I will derive an equation to obtain the current & potential drop across the bulb ...


2

A capacitor is used to store energy in form of electric fields. This electric field is created by charges on plates of capacitor. So, basically you are storing charge on capacitors. Let someone ask you how much charge you can store in your capacitor.What would you reply? Clearly , you reply " I may store 1mC or 100mC, depending on Potential difference ...


2

Georg Ohm's original experiments, 1825, established that for a set temperature, the current through a specific length of a conductor was proportional to the potential difference applied. Ohm's law is empirical; it cannot be derived directly from Maxwell's equations as it depends upon material properties. It is violated by many materials, and even then ...


2

Why are they considering a phase difference of $\phi$? Your calculations are not totally correct. The voltages across different impedants $V_C,V_R,V_L$ have a phase relationship between them and hence the different impedances $Z_C,Z_R,Z_L$ are not directly linearly related as you have done. Consider the following phasor diagram: I hope it is clear ...


2

I understand that capacitance is the ability of a body to store an electrical charge and the formula is $C = {Q \over V}$ Perhaps you just need to top thinking of capacitance as that. "Capacitance" sounds like "capacity", which leads to an intuitive trap like this: If I have a basket with a capacity of 2 apples, then a basket with more capacity can ...


2

Current only flows if there is a voltage across the resistors. If the resistors are ideally short circuited, no current can flow through them.


2

I took the liberty of redrawing this circuit for clarity. Can you take it from here?


2

If you send $R$ to infinity that's equivalent to an open circuit. Since an open circuit, as you can tell from the name, is not a closed loop, Kirchoff's law is not valid, hence your "paradox". Edit: Kirchoff's law is valid, but not in the form you wrote it. You cannot say that $V_R=5 V$ because when you say that $R \to \infty$ what you are saying is that it ...


2

But I wanted to know whether we can charge a capacitor while it is in use If, by "while it is in use", you mean while the capacitor is discharging, i.e., energy is flowing out of the capacitor to some load, then the answer is no since, by definition, if a capacitor is charging, energy is flowing into the capacitor. Put another way, a capacitor cannot ...


2

A lot of these problems are best done by first redrawing the circuit so that it is in a more accessible form. The correct answer is $\frac 8 3 \;\mu$F.


1

Is there an electric field around the poles of the battery before the circuit is attached, and is there still, after the circuit is connected? Yes. However adding the connecting wires is likely to change the distribution of the field. And why is the field equally strong everywhere in the wire, no matter the shape? Usually we design our circuits ...


1

in series the current is the same through each resistor. Not just the same, the current through each is identical. So that the lightbulbs will all have the same brightness but dimmer than if there was just one bulb on there. Well of course, series connected resistances add and so, the total resistance of two series connected bulbs is greater ...


1

Light bulbs, or any loads, in series will all have the same current. This is unrelated to Ohm's Law - it's Kirchhoff's Current Law and it applies if the loads are ohmic or not. Assuming your source voltage stays the same, adding bulbs in series will increase the total resistance which will decrease the total current and make all the bulbs dimmer. The ...


1

I can't understand at all everything after "As the resistance increases ...". Is that really how it was explained? Nonetheless there's an interesting point here. The analysis is not exactly the same as for an ohmic resistor, but not for the reason you suggest. The thing to consider is that the resistance of the light bulb depends on the temperature ...


1

You are correct that light bulbs are non-ohmic (they don't obey Ohm's Law). But that makes no difference. The same current flows through each, even if they have completely different resistances. Electrical current (charge per second) is like the flow of a river. If there are no leaks, and no tributaries joining the river, then the volume of water per ...


1

This is a typical electromagnet. Let's refer to your figure and say that $g$ is the width of the gap and $A_c$ the section of the core. If $g << \sqrt A_c$,the magnetic field $\vec B$ inside the core will be approximately the same as the magnetic field $\vec B_0$ in the air gap (this is because in this case the magnetic field lines will stay ...


1

Schrodinger's Cat explains well why only part of the impendance is taken. Why are they considering a phase difference of $\phi$ ... Also, why are they taking modulus of Z Here's an algebraic explanation: For the RLC circuit, we can write the total impedance in a general form, $$Z_T=R + j Z_r,$$ where $Z_r$ is the total reactive impedance of the $L$'s ...



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